# sols8 - SOLUTIONS TO HOMEWORK ASSIGNMENT #8 1. Graph the...

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SOLUTIONS TO HOMEWORK ASSIGNMENT #8 1. Graph the following functions showing all work: (a) f ( x )= x 2 x - 1 . (b) f ( x e - x 2 , -∞ <x< . (c) f ( x xe - x , -∞ . (d) f ( x x 2 e -| x | . Solution: (a) First notice that the function is not deﬁned at x =1 . In fact lim x 1 + f ( x )=+ and lim x 1 - f ( x -∞ .Thu s x = 1 is a vertical asymptote. Also note that f (0)=0 ,f ( x ) > 0 for x> 1and f ( x ) < 0 for x< 1 . Next notice that y = x +1 is a slant asymptote since long division gives x 2 x - 1 = x +1+ 1 x - 1 . In fact this tells us that f ( x ) approaches the line y = x + 1 from above as x →∞ and from below as x →-∞ . Now we do the calculus: 1. f 0 ( x )=1 - ( x - 1) - 2 =0 ⇐⇒ x , 2 . 2. f 0 ( x ) > 0 ⇐⇒ -∞ 0or2 and f 0 ( x ) < 0 0 1or1 2 . 3. f 00 ( x )=2( x - 1) - 3 is never 0. f 00 ( x ) > 0 f 00 ( x ) < 0 1 . It follows that: 1. f ( x ) is increasing for 0o r 2 and decreasing for 0 1o r1 2 . Therefore there is a local maximum at x = 0 and a local minimum at x =2 . The local maximum and minimum values are f (0) = 0 and f (2) = 4 resp. 2. The graph is concave up for 1 and concave down for -∞ 1 . There are no inﬂection points. Now assemble all this information into the graph. See the diagram at the end. (b) f ( x e - x 2 is deﬁned for all x (therefore no vertical asymptotes). Also notice the following: f (0)=1 , f ( x ) > 0 for all x , f ( - x f ( x ) (i.e. the function is even) and so the graph is symmetric about the y -axis, and lim x →±∞ f ( x )=0 . Now we do the calculus: 1. f 0 ( x - 2 xe - x 2 x 0 ( x ) > 0 0and f 0 ( x ) < 0 0 . 1

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2. f 00 ( x )= - 2 e - x 2 +4 x 2 e - x 2 =(4 x 2 - 2) e - x 2 =0 ⇐⇒ x = ± 1 2 .
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## sols8 - SOLUTIONS TO HOMEWORK ASSIGNMENT #8 1. Graph the...

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